(This structure, $\mathbb Z_8$ is generated by $8\equiv 0$ and is friendly with operations $+$ and $\cdot$.)

We have the following set of odd numbers: $\{ 1,3,5,7\}$. Or, rewriting by $5\equiv-3$ and $7\equiv -1$, this is only
$$\{ 1,3,-3,-1\}$$
The squares of these:
$$(\pm 1)^2 = 1\ \text{ and }\ (\pm 3)^2=9\equiv 1$$

This is a nice example of a direct proof. You start with the facts that if $\phi$ is your positive odd integer, then it is in the form $\phi = 2n + 1$ where $n$ is an integer and $\phi^2 - 1 = 8p \quad( p\in\mathbb Z) $
is true.

Recall that if a number is divisible by 8, then 8 is one of its factors.

This is something like a “case-by-case” proof. You prove the hypothesis for $n \in2\mathbb Z$ and then $n \in 2\mathbb Z + 1$.

Q.E.D

If $n$ is odd, then $n=8q+r$ with $r\in \{ 1,3,5,7 \}$. Then $n^2=64q^2+16qr+r^2=8(8q^2+2r)+r^2$. Thus, it suffices to prove that $r^2-1$ is divisible by $8$, which can be done by a simple calculation.

Note that $n^2 - 1 = (n+1)(n-1)$. Because n is odd, both $n+1$ and $n-1$ are even. Let $m=n-1$. Noting that $2|m$, consider both $m$ and $m+2$ mod 4. After we see that either $m$ or $m+2$ is divisible by 4, we know that $m(m+2) = (2k)(4l) = 8kl$. (WLOG due the the commutativity of multiplication)